Formula & theorem reference

This is nothing but the unit-circle definitions of cosine and sine — \cos\theta and \sin\theta are the x- and y-coordinates on a circle of radius 1 — stretched from radius 1 out to radius r.

r = \sqrt{x^2+y^2} is just the distance formula from the origin to (x,y). Dividing the two conversion equations above, \frac{y}{x} = \frac{r\sin\theta}{r\cos\theta} = \tan\theta, whenever x \neq 0.

Approximate the curve by a polygonal path connecting points at x_0,x_1,\ldots,x_n. Between consecutive points, the path rises \Delta y_i over a run of \Delta x, so that segment's length is \sqrt{(\Delta x)^2+(\Delta y_i)^2} = \sqrt{1+(\Delta y_i/\Delta x)^2}\,\Delta x. By the Mean Value Theorem, \Delta y_i/\Delta x = f'(x_i^*) for some point x_i^* in that subinterval, so the segment's length is \sqrt{1+[f'(x_i^*)]^2}\,\Delta x. Summing over all n segments gives a Riemann sum for \sqrt{1+[f'(x)]^2}, and letting n\to\infty turns that sum into the stated integral, with the polygonal path's total length converging to the true arc length.

Start from the product rule, \dfrac{d}{dx}[uv] = u\,\dfrac{dv}{dx} + v\,\dfrac{du}{dx}, and integrate both sides with respect to x: uv = \int u\,\frac{dv}{dx}\,dx + \int v\,\frac{du}{dx}\,dx = \int u\,dv + \int v\,du. Solving for \int u\,dv gives the formula directly. No information is added or lost — it's the product rule read from right to left, with the two terms it produces separated onto opposite sides of the equation.

Fix x in (a,b) and form the difference quotient defining A'(x): \frac{A(x+h)-A(x)}{h} = \frac{1}{h}\left[\int_a^{x+h}f(t)\,dt -\int_a^x f(t)\,dt\right] = \frac{1}{h}\int_x^{x+h} f(t)\,dt, the last step by additivity of the definite integral over adjacent intervals. The remaining integral is f's net signed area over a sliver of width h -- and by the Mean Value Theorem for Integrals (f continuous on [x,x+h]), that sliver equals f(c)\cdot h for SOME c between x and x+h, so \frac{A(x+h)-A(x)}{h} = f(c) \quad\text{for some $c$ strictly between $x$ and $x+h$.} As h\to0, c is squeezed between x and x+h and is forced to x too, and f's continuity at x then forces f(c)\to f(x). Taking the limit, A'(x) = \lim_{h\to0}\frac{A(x+h)-A(x)}{h} = \lim_{h\to0} f(c) = f(x).

The same conclusion falls out of the Riemann-sum definition directly, without invoking the Mean Value Theorem for Integrals: for small h, \int_x^{x+h} f(t)\,dt is itself approximated by a single thin rectangle of width h and height approximately f(x) (since f barely changes over an interval that short, by continuity), so \frac{1}{h}\int_x^{x+h}f(t)\,dt\approx f(x), with the approximation becoming exact in the limit h\to0 -- the same "thin slice, height \approx f(x)" idea Chapter 7 built Riemann sums from in the first place, just run over a single infinitesimally short interval instead of the whole [a,b].

Continuity of f is load-bearing in both arguments: it is what supplies the Mean Value Theorem for Integrals' own hypothesis, and it is what forces f(c)\to f(x) as the squeezed point c\to x -- a discontinuous f can make A fail to be differentiable exactly at the discontinuity, even though A itself (an accumulated area) still exists there.

Let Q=(x_2,y_1,z_1) and R=(x_2,y_2,z_1) sit on the path from P_1 to P_2 that moves along one axis at a time. In the horizontal plane z=z_1, P_1, Q, R form a right triangle with legs |x_2-x_1| and |y_2-y_1|, so by the Pythagorean theorem

Now R and P_2 differ only in z, and the segment RP_2 is perpendicular to the plane z=z_1 — hence perpendicular to P_1R — so P_1, R, P_2 form a second right triangle, this one with legs d(P_1,R) and |z_2-z_1|:

Taking the positive square root gives the formula.

Draw \mathbf{u} and \mathbf{v} from a common point O, with tips at U and V. The triangle OUV has two sides |\mathbf{u}|, |\mathbf{v}| meeting at angle \theta, and third side |\mathbf{u}-\mathbf{v}|, so the Law of Cosines gives

Expanding the left side component-by-component, \mathbf{u}-\mathbf{v}=\langle u_1-v_1,\,u_2-v_2,\,u_3-v_3\rangle, so

Both expressions equal |\mathbf{u}-\mathbf{v}|^2; setting them equal, canceling |\mathbf{u}|^2+|\mathbf{v}|^2 from both sides, and dividing by -2 leaves \sum_i u_iv_i = |\mathbf{u}||\mathbf{v}|\cos\theta, which is exactly \mathbf{u}\cdot\mathbf{v}=u_1v_1+u_2v_2+u_3v_3 on the left.

Drop a perpendicular from the tip of \mathbf{v} down onto the line through \mathbf{u}; this forms a right triangle with hypotenuse |\mathbf{v}| and angle \theta between \mathbf{u} and \mathbf{v}. Exactly as in Chapter 8, the leg lying along \mathbf{u}'s line — the scalar projection — has length |\mathbf{v}|\cos\theta (adjacent equals hypotenuse times cosine). From the theorem above, \cos\theta = \frac{\mathbf{u}\cdot\mathbf{v}}{|\mathbf{u}|\,|\mathbf{v}|}, so the scalar projection simplifies to

The vector projection points along \mathbf{u}'s own direction, so multiply this scalar length by the unit vector \frac{\mathbf{u}}{|\mathbf{u}|}:

Direct computation confirms perpendicularity: expanding the dot product \mathbf{u}\cdot(\mathbf{u}\times\mathbf{v}) component-by-component gives u_1(u_2v_3-u_3v_2)+u_2(u_3v_1-u_1v_3)+u_3(u_1v_2-u_2v_1), and every term cancels against another, leaving 0; the same cancellation shows \mathbf{v}\cdot(\mathbf{u}\times\mathbf{v})=0.

For the magnitude, Lagrange's identity states |\mathbf{u}\times\mathbf{v}|^2 = |\mathbf{u}|^2|\mathbf{v}|^2 - (\mathbf{u}\cdot\mathbf{v})^2 (both sides expand to the same sum of squared component products — a direct algebraic check). Substituting \mathbf{u}\cdot\mathbf{v}=|\mathbf{u}||\mathbf{v}|\cos\theta:

Since 0\le\theta\le\pi makes \sin\theta\ge0, taking square roots gives |\mathbf{u}\times\mathbf{v}|=|\mathbf{u}||\mathbf{v}|\sin\theta.

\mathbf{v}\times\mathbf{w} is perpendicular to the base parallelogram spanned by \mathbf{v} and \mathbf{w}, with length equal to that base's area. Writing \mathbf{n}=\mathbf{v}\times\mathbf{w} and \hat{\mathbf{n}}=\mathbf{n}/|\mathbf{n}|, the parallelepiped's height — how far \mathbf{u} reaches in the direction perpendicular to the base — is the absolute value of the scalar projection of \mathbf{u} onto \hat{\mathbf{n}}, namely |\mathbf{u}\cdot\hat{\mathbf{n}}|. Volume is base area times height:

The determinant form follows from expanding \mathbf{v}\times\mathbf{w} into components and dotting with \mathbf{u} — exactly the cofactor expansion of the 3\times3 determinant along its first row.

A point Q=(x,y,z) lies on the line through P_0 in direction \mathbf{v} exactly when \overrightarrow{P_0Q} is parallel to \mathbf{v} — that is, \overrightarrow{P_0Q}=t\mathbf{v} for some real t. Writing \mathbf{r}=\langle x,y,z\rangle and \mathbf{r}_0=\langle x_0,y_0,z_0\rangle, this parallelism condition is exactly \mathbf{r}-\mathbf{r}_0=t\mathbf{v}, i.e. \mathbf{r}=\mathbf{r}_0+t\mathbf{v}. Reading off each component gives the parametric form; eliminating t from each of t=\frac{x-x_0}{a}=\frac{y-y_0}{b}=\frac{z-z_0}{c} (valid whenever a,b,c\neq0) gives the symmetric form.

A point Q=(x,y,z) lies in the plane exactly when \overrightarrow{P_0Q}=\langle x-x_0,\,y-y_0,\,z-z_0\rangle lies flat in the plane itself, which is exactly the condition that \overrightarrow{P_0Q} is perpendicular to the normal \mathbf{n}: \mathbf{n}\cdot\overrightarrow{P_0Q}=0. Writing that dot product out in components gives a(x-x_0)+b(y-y_0)+c(z-z_0)=0 directly, and collecting the constant terms on the right gives ax+by+cz=ax_0+by_0+cz_0=d.

Let P_0=(x_0,y_0,z_0) be any point that IS on the plane, so ax_0+by_0+cz_0=d. The distance D is the length of whatever piece of \overrightarrow{P_0P_1} points straight off the plane — exactly the absolute value of the scalar projection of \overrightarrow{P_0P_1} onto the unit normal \hat{\mathbf{n}}=\mathbf{n}/|\mathbf{n}|:

Substituting ax_0+by_0+cz_0=d gives D=\dfrac{|ax_1+by_1+cz_1-d|}{|\mathbf{n}|}, which no longer references P_0 at all — any point on the plane gives the same answer, as it must.

Write the difference quotient component by component: \frac{\mathbf{r}(t+\Delta t) - \mathbf{r}(t)}{\Delta t} = \left\langle \frac{f(t+\Delta t)-f(t)}{\Delta t}, \frac{g(t+\Delta t)-g(t)}{\Delta t}, \frac{h(t+\Delta t)-h(t)}{\Delta t}\right\rangle. A vector-valued limit is taken component by component (each component is an ordinary real-valued limit), so letting \Delta t \to 0 sends each bracketed quotient to f'(t), g'(t), h'(t) respectively, giving \mathbf{r}'(t) = \langle f'(t), g'(t), h'(t)\rangle.

Each rule reduces to the ordinary single-variable product/chain rule applied component by component. For the dot-product rule, write \mathbf{u}\cdot\mathbf{v} = u_1v_1+u_2v_2+u_3v_3 and differentiate each term with the scalar product rule: \frac{d}{dt}(u_iv_i) = u_i'v_i + u_iv_i'. Summing the three terms regroups exactly into $\mathbf{u}'\cdot\mathbf{v}

• \mathbf{u}\cdot\mathbf{v}'. The cross-product rule follows the same way, differentiating the determinant expansion of \mathbf{u}\times \mathbf{v}$ term by term and keeping every factor's original left/right position, since each entry of the resulting vector is itself a sum of products of components.

Antidifferentiation undoes differentiation, and \mathbf{r}'(t)=\langle f'(t),g'(t),h'(t)\rangle already shows a vector-valued derivative acts component by component. Running that statement backward, an antiderivative of \mathbf{r}(t) must have f(t), g(t), h(t) as the derivatives of its own three components, so each component of \int\mathbf{r}(t)\,dt is exactly the ordinary antiderivative of the matching component of \mathbf{r}(t). Each component contributes its own scalar constant, and the three scalar constants bundle into one constant vector \mathbf{C}.

Partition [a,b] into n pieces of width \Delta t. The straight-line distance between consecutive sample points \mathbf{r}(t_{i-1}) and \mathbf{r}(t_i) approximates |\mathbf{r}'(t_i)|\,\Delta t, the same "speed times a short time interval" approximation used for a plane curve, since |\mathbf{r}'(t)| is exactly the instantaneous speed \dfrac{ds}{dt}. Summing these n straight-line pieces gives a Riemann sum \sum |\mathbf{r}'(t_i)|\,\Delta t for the length of the inscribed polygonal path; letting n \to \infty turns the sum into \int_a^b |\mathbf{r}'(t)|\,dt, and the polygonal path's length converges to the curve's own length because \mathbf{r} is smooth (continuously differentiable with \mathbf{r}'(t)\ne\mathbf{0}).

Since \mathbf{T}(t)\cdot\mathbf{T}(t)=1 for every t (it's a unit vector), differentiating both sides with the dot-product rule gives 2\,\mathbf{T}(t)\cdot\mathbf{T}'(t) = 0, so \mathbf{T}'(t) is always perpendicular to \mathbf{T}(t) — that perpendicularity is exactly what makes \mathbf{N}(t), the normalized \mathbf{T}'(t), well-defined as "the direction the tangent is turning toward." For the cross-product form: \mathbf{r}'=|\mathbf{r}'|\mathbf{T}, so differentiating with the product rule gives \mathbf{r}'' = |\mathbf{r}'|'\mathbf{T} + |\mathbf{r}'|\mathbf{T}'. Crossing both sides with \mathbf{r}' kills the first term (a vector crossed with a parallel multiple of itself is \mathbf{0}), leaving \mathbf{r}'\times\mathbf{r}'' = |\mathbf{r}'|^2(\mathbf{T}\times\mathbf{T}'), and since \mathbf{T} and \mathbf{T}' are perpendicular, |\mathbf{T}\times\mathbf{T}'| = |\mathbf{T}'|. Solving for |\mathbf{T}'| and substituting into \kappa = |\mathbf{T}'|/|\mathbf{r}'| produces the stated formula.

\mathbf{T}\cdot\mathbf{T}=1, \mathbf{N}\cdot\mathbf{N}=1, and \mathbf{T}\cdot\mathbf{N}=0 hold at every s; differentiating each with the product rule shows d\mathbf{T}/ds is perpendicular to \mathbf{T} (so it's some combination of \mathbf{N} and \mathbf{B}) and, by curvature's own definition \kappa=|d\mathbf{T}/ds| together with \mathbf{N}'s definition as the normalized d\mathbf{T}/ds, that combination is exactly \kappa\mathbf{N} with no \mathbf{B} part. Differentiating \mathbf{B}=\mathbf{T}\times\mathbf{N} and substituting d\mathbf{T}/ds=\kappa\mathbf{N} kills the first term (a vector crossed with a parallel multiple of itself), leaving d\mathbf{B}/ds = \mathbf{T}\times(d\mathbf{N}/ds), which turns out to have no \mathbf{T}-component and no \mathbf{B}-component (both follow from differentiating \mathbf{B}\cdot\mathbf{T}=0 and \mathbf{B}\cdot\mathbf{B}=1 the same way) — so d\mathbf{B}/ds is a multiple of \mathbf{N} alone; that multiple is DEFINED to be -\tau. Finally \mathbf{N}=\mathbf{B}\times\mathbf{T}, so differentiating and substituting the first two formulas produces the middle one, d\mathbf{N}/ds = -\kappa\mathbf{T}+\tau\mathbf{B}.

Write \mathbf{r}' = v\mathbf{T} (velocity is speed times direction) and differentiate with the product rule: \mathbf{a} = \mathbf{r}'' = v'\mathbf{T} + v\mathbf{T}'. Since \mathbf{T}' = |\mathbf{T}'|\mathbf{N} by the definition of \mathbf{N}, and \kappa = |\mathbf{T}'|/v gives |\mathbf{T}'| = \kappa v, the second term becomes v(\kappa v)\mathbf{N} = \kappa v^2\mathbf{N}. So \mathbf{a} = v'\mathbf{T} + \kappa v^2\mathbf{N} — no \mathbf{B} term appears at all, matching a_T=v', a_N=\kappa v^2. The dot/cross-product forms of a_T, a_N follow by dotting or crossing \mathbf{a} = v'\mathbf{T}+\kappa v^2\mathbf{N} with \mathbf{r}'=v\mathbf{T}: dotting kills the \mathbf{N} term (\mathbf{T}\cdot\mathbf{N}=0) and leaves v v', matching \mathbf{r}'\cdot\mathbf{r}''=vv' divided by v; crossing kills the \mathbf{T} term (\mathbf{T}\times\mathbf{T}=\mathbf{0}) and leaves v\cdot\kappa v^2(\mathbf{T}\times\mathbf{N}), a unit vector magnitude \kappa v^3, matching |\mathbf{r}'\times\mathbf{r}''| divided by v.

Fix (a,b) and consider the second difference \Delta(h,k) = f(a+h,b+k) - f(a+h,b) - f(a,b+k) + f(a,b). Define g(x) = f(x,b+k) - f(x,b), so \Delta(h,k) = g(a+h)-g(a). The Mean Value Theorem gives \Delta(h,k) = h\,g'(c) = h\big(f_x(c,b+k) - f_x(c,b)\big) for some c between a and a+h, and applying the Mean Value Theorem again to f_x(c,\cdot) gives \Delta(h,k) = hk\,f_{xy}(c,d) for some d between b and b+k. Running the same argument the other order (differencing in y first, then x) gives \Delta(h,k) = hk\,f_{yx}(c',d') for some other point (c',d'). Both (c,d) and (c',d') approach (a,b) as (h,k)\to(0,0), so continuity of f_{xy} and f_{yx} at (a,b) forces f_{xy}(a,b) = \lim_{(h,k)\to(0,0)} \frac{\Delta(h,k)}{hk} = f_{yx}(a,b).

Let g(t) = f(x_0+ta,\,y_0+tb), so D_{\mathbf{u}}f(x_0,y_0) = g'(0) by the limit definition above. g is f composed with the line x(t)=x_0+ta,\ y(t)=y_0+tb, so the chain rule gives g'(t) = f_x(x(t),y(t))\,x'(t) + f_y(x(t),y(t))\,y'(t) = f_x(x(t),y(t))\,a + f_y(x(t),y(t))\,b. Evaluating at t=0 gives g'(0) = f_x(x_0,y_0)\,a + f_y(x_0,y_0)\,b.

Write \mathbf{u}=\langle a,b\rangle. The directional-derivative formula already gives D_{\mathbf{u}}f = f_x a + f_y b, and \langle f_x,f_y\rangle\cdot\langle a,b\rangle = f_x a + f_y b by the definition of the dot product — the two expressions are the same number, just written two different ways.

Let \theta be the angle between \mathbf{u} and \nabla f(x_0,y_0). Since \mathbf{u} is a unit vector, the dot product formula for D_{\mathbf{u}}f = \nabla f\cdot\mathbf{u} becomes D_{\mathbf{u}}f(x_0,y_0) = |\nabla f(x_0,y_0)|\,|\mathbf{u}|\cos\theta = |\nabla f(x_0,y_0)|\cos\theta. Since -1\le\cos\theta\le 1, this quantity ranges over exactly [-|\nabla f(x_0,y_0)|,\,|\nabla f(x_0,y_0)|], hit at \theta=0 (u points along \nabla f) and \theta=\pi (u points opposite it) respectively.

Parametrize the level curve near (x_0,y_0) as \mathbf{r}(t) = \langle x(t), y(t)\rangle with \mathbf{r}(0) = \langle x_0,y_0\rangle and f(x(t),y(t)) = k for every t near 0. Differentiating both sides with respect to t (chain rule) gives f_x(x(t),y(t))\,x'(t) + f_y(x(t),y(t))\,y'(t) = 0, which is exactly \nabla f(x(t),y(t))\cdot\mathbf{r}'(t) = 0. At t=0, \mathbf{r}'(0) is tangent to the level curve, so \nabla f(x_0,y_0)\cdot\mathbf{r}'(0) = 0 says the gradient is orthogonal to that tangent direction.

Let \mathbf{r}(t) = \langle x(t),y(t),z(t)\rangle be ANY curve lying on the surface with \mathbf{r}(0) = \langle x_0,y_0,z_0\rangle, so F(x(t),y(t),z(t)) = k for every t near 0. The chain rule gives F_x\,x'(t) + F_y\,y'(t) + F_z\,z'(t) = 0, i.e. \nabla F(\mathbf{r}(t))\cdot\mathbf{r}'(t) = 0. At t=0 this says \nabla F(x_0,y_0,z_0) is orthogonal to \mathbf{r}'(0) — the tangent vector of an arbitrary curve on the surface through the point. Since the curve was arbitrary, \nabla F(x_0,y_0,z_0) is orthogonal to every tangent vector to the surface there, which is what "normal to the surface" means.

Near (a,b), Taylor's theorem in two variables writes f as its constant value plus a quadratic form built from the second partials (the linear terms vanish because (a,b) is critical): f(a+h, b+k) \approx f(a,b) + \tfrac12\left(f_{xx}h^2 + 2f_{xy}hk + f_{yy}k^2\right). Completing the square in h (treating f_{xx} \neq 0) turns the bracket into f_{xx}\left(h + \frac{f_{xy}}{f_{xx}}k\right)^2 + \frac{f_{xx}f_{yy} - f_{xy}^2}{f_{xx}}k^2 = f_{xx}\left(h + \frac{f_{xy}}{f_{xx}}k\right)^2 + \frac{D}{f_{xx}}k^2. When D>0, both squared terms carry the same sign as f_{xx} for every direction (h,k), so the quadratic form never changes sign near (a,b) — a min if that sign is positive, a max if negative. When D<0, the two terms have opposite signs: moving purely in h (with k=0) gives one sign, moving purely in k-dominated directions gives the other, so f rises in some directions and falls in others — the saddle signature. When D=0 the quadratic approximation degenerates along some direction, and the cubic-and-higher terms Taylor's theorem dropped are what actually decide the case, which is exactly why the test goes silent there.

The single-variable Extreme Value Theorem needs a closed, bounded interval; this is the exact same statement one dimension up, and both hypotheses are load-bearing in the same way. "Bounded" keeps D from letting f's values run off to infinity along some escaping direction. "Closed" — D contains its own boundary curve, not just its interior — keeps a candidate maximum from sitting at a boundary point that's excluded from the region, the way f(x,y) = x on the open disk x^2+y^2<1 has no maximum: any point falls short of the open edge, and a point arbitrarily close to the edge is always beaten by one closer still. A full proof again goes through compactness (a closed, bounded region in \mathbb{R}^n is exactly a compact one) and a convergent-subsequence argument identical in spirit to the one-variable case; what matters here is the payoff — on a closed, bounded region, the candidates test (interior critical points plus the entire boundary) is guaranteed to contain the winner.

Parametrize the constraint curve near (a,b) by arc length, \mathbf{r}(t) = (x(t), y(t)), with \mathbf{r}(0) = (a,b) — this is possible precisely because \nabla g(a,b) \neq \mathbf{0}, which keeps the level curve g=k smooth with a well-defined tangent there. Restrict f to the curve: h(t) = f(x(t), y(t)). If f has a local extremum at (a,b) subject to the constraint, then h has an ordinary single-variable local extremum at t=0, so h'(0) = 0. The chain rule expands that as h'(0) = f_x(a,b)\,x'(0) + f_y(a,b)\,y'(0) = \nabla f(a,b)\cdot \mathbf{r}'(0) = 0, meaning \nabla f(a,b) is perpendicular to the curve's tangent vector \mathbf{r}'(0) at that point. But \nabla g(a,b) is also perpendicular to the same tangent vector — that's what it means for g=k to be a level curve of g. Two vectors both perpendicular to the same tangent line in the plane must be parallel to each other, so \nabla f(a,b) = \lambda \nabla g(a,b) for some scalar \lambda.

The intersection of the two surfaces g=k_1 and h=k_2 is (near \mathbf{p}, given the non-parallel gradient hypothesis) a smooth space curve, tangent line \mathbf{r}'(0) at \mathbf{p}. The same restrict-and-differentiate argument as the single-constraint case applies: if f has a constrained extremum at \mathbf{p}, then \nabla f(\mathbf{p}) is perpendicular to \mathbf{r}'(0). So are \nabla g(\mathbf{p}) and \nabla h(\mathbf{p}), since the curve lies in both level surfaces at once. The set of vectors perpendicular to a single line in \mathbb{R}^3 is an entire plane, not just a single direction — so unlike the one-constraint case, being perpendicular to \mathbf{r}'(0) doesn't force \nabla f to be parallel to any one gradient. It does force \nabla f(\mathbf{p}) to lie in the plane that \nabla g(\mathbf{p}) and \nabla h(\mathbf{p}) themselves span (the normal plane to the curve, since both are perpendicular to \mathbf{r}'(0) and non-parallel to each other) — and every vector in that plane is some combination \lambda \nabla g + \mu \nabla h.

Fix x and let A(x) = \int_c^d f(x,y)\,dy, the (signed) area of the cross-section cut by the vertical plane at that x. Slicing the solid region under z=f(x,y) into thin slabs of thickness dx, each slab has volume approximately A(x)\,dx, and summing all the slabs gives \int_a^b A(x)\,dx as the total volume — the definition of the double integral itself. This is exactly the cross-sectional-area argument behind volume by slicing from single-variable calculus, run in the x-direction instead of a fixed axis. The same slicing argument in the y-direction produces the other iterated order, and continuity of f on a closed rectangle guarantees both limits exist and agree.

Partition the region with n rays and m concentric arcs. One polar "grid cell" between radii r_i and r_i+\Delta r and angles \theta_j and \theta_j+\Delta\theta is an annular sector, whose exact area is \frac12(r_i+\Delta r)^2\Delta\theta - \frac12 r_i^2\Delta\theta = \left(r_i + \frac{\Delta r}{2}\right)\Delta r\,\Delta\theta. As \Delta r \to 0, this area is r_i\,\Delta r\,\Delta\theta + O(\Delta r^2), so the Riemann sum \sum f(r_i\cos\theta_j, r_i\sin\theta_j)\, r_i\,\Delta r\,\Delta\theta converges to the double integral with the extra factor of r built in, exactly the substitution x=r\cos\theta, y=r\sin\theta carries into the area element.

Partition R into small rectangles of dimensions \Delta x \times \Delta y. Over one such piece, the surface is approximated by a parallelogram cut from the tangent plane at that point, spanned by the vectors \mathbf{a}=\langle \Delta x, 0, f_x\Delta x\rangle (walking \Delta x in the x-direction along the tangent plane) and \mathbf{b}=\langle 0, \Delta y, f_y\Delta y\rangle. That parallelogram's area is |\mathbf{a}\times\mathbf{b}| = \sqrt{f_x^2+f_y^2+1}\;\Delta x\,\Delta y, so summing every piece and refining the partition turns the sum into \iint_R \sqrt{1+f_x^2+f_y^2}\,dA — the same tangent-plane-approximation idea behind arc length in one variable, extended by one dimension.

Slice E by fixing (x,y)\in D and integrating f over the segment of the vertical line through (x,y) that lies inside E — that inner integral is exactly \int_{u_1(x,y)}^{u_2(x,y)} f(x,y,z)\,dz, a function of x and y alone. Fubini's theorem for double integrals then lets that result be integrated over D in either x-then-y or y-then-x order, since D is itself an ordinary plane region. The Type II and Type III versions follow by relabeling which pair of variables describes the shadow and which variable the two surfaces bound.

Build the box [\rho,\rho+d\rho]\times[\theta,\theta+d\theta]\times[\phi,\phi+d\phi] one edge at a time. Moving \rho by d\rho with \theta,\phi fixed sweeps a straight radial edge of length d\rho. Moving \phi by d\phi with \rho,\theta fixed sweeps an arc of a great circle of radius \rho, length \rho\,d\phi. Moving \theta by d\theta with \rho,\phi fixed sweeps an arc around the z-axis at horizontal radius \rho\sin\phi (the same r a cylindrical slice at that height would have), length \rho\sin\phi\,d\theta. The three edges are mutually perpendicular, so the box's volume is their product: d\rho\cdot(\rho\,d\phi)\cdot(\rho\sin\phi\,d\theta) = \rho^2\sin\phi\,d\rho\,d\phi\,d\theta.

Near (u_0,v_0), the transformation's best linear approximation sends the edge vector (du,0) to \left(\frac{\partial x}{\partial u}, \frac{\partial y}{\partial u}\right)du and the edge vector (0,dv) to \left(\frac{\partial x}{\partial v}, \frac{\partial y}{\partial v}\right)dv — the two columns of the Jacobian matrix, scaled. The image of the small du\times dv rectangle is (to first order) the parallelogram spanned by those two vectors, and the area of a parallelogram spanned by (a,b) and (c,d) is |ad-bc|, exactly the absolute value of that 2\times2 determinant.

Partition S into small rectangles of area du\,dv. Each maps, to first order, onto a parallelogram in R of area \left|\frac{\partial(x,y)}{\partial(u,v)}\right|du\,dv (the Jacobian's own defining property). Summing f times the image area over every small piece and refining the partition gives a Riemann sum for the left side built out of a Riemann sum for the right side; the two limits agree because they are both approximating the same collection of shrinking pieces, just measured in different coordinates.

By the chain rule, \dfrac{d}{dt}f(\mathbf{r}(t)) = \nabla f(\mathbf{r}(t)) \cdot \mathbf{r}'(t). So \int_C \nabla f \cdot d\mathbf{r} = \int_a^b \nabla f(\mathbf{r}(t)) \cdot \mathbf{r}'(t)\, dt = \int_a^b \frac{d}{dt}f(\mathbf{r}(t))\, dt = f(\mathbf{r}(b)) - f(\mathbf{r}(a)) = f(B) - f(A), the last step being the ordinary single-variable Fundamental Theorem of Calculus applied to the scalar function t \mapsto f(\mathbf{r}(t)).

(1 \Rightarrow 2) is the Fundamental Theorem for Line Integrals: if \mathbf{F} = \nabla f, every curve from A to B gives f(B) - f(A), regardless of the route taken.

(2 \Rightarrow 3): a closed curve C starts and ends at the same point A, so under path independence its integral must equal the integral over the degenerate curve that never leaves A, namely 0.

(3 \Rightarrow 2): given two curves C_1, C_2 from A to B, let C be C_1 followed by C_2 reversed — a closed loop. Then 0 = \oint_C \mathbf{F}\cdot d\mathbf{r} = \int_{C_1}\mathbf{F}\cdot d\mathbf{r} - \int_{C_2}\mathbf{F}\cdot d\mathbf{r}, so the two path integrals agree.

(2 \Rightarrow 1): fix a basepoint A_0 \in D and define f(A) = \int_{A_0}^{A}\mathbf{F}\cdot d\mathbf{r} along any path (well defined by path independence). A direct computation shows \nabla f = \mathbf{F}.

(\Rightarrow) If \mathbf{F} = \nabla f, then P = f_x and Q = f_y, so P_y = f_{xy} and Q_x = f_{yx}. Because P and Q have continuous partials, f's second partials are continuous too, and Clairaut's Theorem gives f_{xy} = f_{yx}, hence P_y = Q_x.

(\Leftarrow) The converse needs simple connectedness: it is exactly the hypothesis Green's Theorem requires to convert \oint_C \mathbf{F}\cdot d\mathbf{r} over any closed curve C in D into \iint_R (Q_x - P_y)\, dA = 0, which forces every closed-loop integral to vanish — the third equivalent condition for conservative fields. On a region with a hole (where D is not simply connected), P_y = Q_x can hold everywhere \mathbf{F} is defined without \mathbf{F} being conservative, since a loop around the hole isn't the boundary of a region entirely inside D.

Sketch for a region R that is both x-simple and y-simple (the general case patches finitely many such regions together, with the shared interior edges canceling since each is traversed once in each direction). Write R = \{(x,y) : a \le x \le b,\ g_1(x) \le y \le g_2(x)\}. Then \iint_R -\frac{\partial P}{\partial y}\, dA = -\int_a^b \big[P(x,g_2(x)) - P(x,g_1(x))\big]\, dx, by the single-variable Fundamental Theorem applied to the inner integral. The right side is exactly \oint_C P\,dx, splitting C into its bottom piece y=g_1(x) (traversed left to right) and top piece y=g_2(x) (traversed right to left, contributing the minus sign). The same argument with R described as y-simple gives \iint_R \frac{\partial Q}{\partial x}\, dA = \oint_C Q\,dy. Adding the two established identities gives the theorem.

Hold v=v_0 fixed and let u vary: \mathbf{r}(u,v_0) traces a curve lying entirely on the surface and passing through \mathbf{r}(u_0,v_0), with velocity vector \mathbf{r}_u(u_0,v_0) — so \mathbf{r}_u is tangent to the surface there. The same argument with u=u_0 fixed shows \mathbf{r}_v is tangent to the surface too, along the perpendicular grid direction. Two non-parallel vectors tangent to the surface at the same point span its entire tangent plane, and the cross product of two vectors spanning a plane is normal to that plane — so \mathbf{r}_u\times\mathbf{r}_v is a normal vector, and the tangent plane is the plane through \mathbf{r}(u_0,v_0) with that normal.

Partition D into small rectangles of dimensions \Delta u\times\Delta v. A single such rectangle, based at (u_0,v_0), maps under \mathbf{r} to a small curved patch that's well approximated by the parallelogram spanned by \mathbf{r}_u(u_0,v_0)\,\Delta u and \mathbf{r}_v(u_0,v_0)\,\Delta v — the same tangent-plane linearization argument behind every "dS from two tangent vectors" formula in this course. That parallelogram's area is |\mathbf{r}_u\Delta u \times \mathbf{r}_v\Delta v| = |\mathbf{r}_u\times\mathbf{r}_v|\;\Delta u\,\Delta v. Summing this over every rectangle in the partition and refining the partition turns the Riemann sum into \iint_D |\mathbf{r}_u\times \mathbf{r}_v|\,du\,dv.

Partition D into small rectangles as in the surface-area formula. Each rectangle's image patch has area \approx |\mathbf{r}_u\times\mathbf{r}_v|\,\Delta u\,\Delta v; weighting that patch by f evaluated at the corresponding point and summing over every patch gives a Riemann sum \sum f(\mathbf{r}(u_i,v_i))\,|\mathbf{r}_u\times\mathbf{r}_v|\, \Delta u\,\Delta v. Refining the partition turns the sum into the double integral above — exactly the same "weight the area element by f" move that turns arc length \int_C ds into the scalar line integral \int_C f\,ds, one dimension up.

Write d\mathbf{S} = \mathbf{n}\,dS for the unit normal \mathbf{n} = \dfrac{\mathbf{r}_u\times\mathbf{r}_v}{|\mathbf{r}_u\times\mathbf{r}_v|} in the chosen orientation. By the scalar surface integral, dS = |\mathbf{r}_u\times\mathbf{r}_v|\,du\,dv, so d\mathbf{S} = \frac{\mathbf{r}_u\times\mathbf{r}_v} {|\mathbf{r}_u\times\mathbf{r}_v|}\,|\mathbf{r}_u\times\mathbf{r}_v|\, du\,dv = (\mathbf{r}_u\times\mathbf{r}_v)\,du\,dv — the two magnitude factors cancel, leaving the bare cross product as the vector area element. Dotting \mathbf{F} against d\mathbf{S} and integrating over D gives the formula above, the direct surface analogue of \int_C \mathbf{F}\cdot d\mathbf{r} = \int_a^b \mathbf{F}(\mathbf{r}(t))\cdot\mathbf{r}'(t)\,dt for a line integral of a vector field.

(\Rightarrow) If \mathbf{F} = \nabla f, then P=f_x, Q=f_y, R=f_z. Each component of \operatorname{curl}\mathbf{F} is a difference of mixed second partials of f — e.g. the \mathbf{k}-component is f_{yx}-f_{xy} — which vanishes by Clairaut's Theorem since f's second partials are continuous. The same argument kills the other two components, so \operatorname{curl}\mathbf{F}=\mathbf{0}.

(\Leftarrow) The converse needs simple connectedness in space: every closed curve in D must be shrinkable to a point without leaving D (a stronger condition than merely "no holes" — a region built around a solid tube, for instance, is not simply connected even though it has no cavity). Given \operatorname{curl}\mathbf{F}=\mathbf{0} on such a D, Stokes' Theorem (later in this chapter) lets any closed curve C in D be spanned by a surface S entirely inside D, giving \oint_C \mathbf{F}\cdot d\mathbf{r} = \iint_S \operatorname{curl}\mathbf{F}\cdot d\mathbf{S} = 0 — zero circulation around every closed loop, which is the closed-loop form of path independence.

Apply the circulation (tangential) form of Green's Theorem to the rotated field \mathbf{G} = Q\,\mathbf{i} - P\,\mathbf{j} — rotating \mathbf{F} by -90° at every point turns "tangent to C" into "normal to C" and vice versa, since rotating \mathbf{T}=(dx/ds,\,dy/ds) by -90° gives exactly the outward normal \mathbf{n}=(dy/ds,\,-dx/ds) for a positively oriented C. Then \mathbf{G}\cdot d\mathbf{r} = \mathbf{F}\cdot\mathbf{n}\,ds, and the circulation form gives \oint_C \mathbf{G}\cdot d\mathbf{r} = \iint_R \Big(\underbrace{(-P)_x - Q_y}_{\text{curl of }\mathbf{G}}\Big)\,dA \cdot(-1)= \iint_R (P_x+Q_y)\,dA = \iint_R \operatorname{div}\mathbf{F}\,dA, matching the flux integral on the left to the divergence integral on the right.

Sketch for S a graph z=g(x,y) over a region R in the xy-plane, with C the boundary of R lifted onto the graph. Parametrize S by (x,y)\mapsto(x,y,g(x,y)), so d\mathbf{S} = (-g_x,-g_y,1)\,dA. Writing out \operatorname{curl}\mathbf{F}\cdot d\mathbf{S} in terms of P, Q, R and their partials, then substituting z=g(x,y) throughout, the right side reduces — after applying the multivariable chain rule to track how P and Q change along the lifted curve — to exactly \oint_{\partial R} \big[P(x,y,g(x,y))+R(x,y,g(x,y))g_x\big]\,dx + \big[Q(x,y,g(x,y))+R(x,y,g(x,y))g_y\big]\,dy, which the circulation form of Green's Theorem converts into a double integral over R identical to the one obtained by substituting z=g(x,y) into the left-hand side's curl-flux integrand directly. A general piecewise-smooth S is patched together from finitely many such graphs, with shared interior edges canceling since each is traversed once in each direction.

Sketch for E a solid that is simultaneously x-simple, y-simple, and z-simple (a general E patches together finitely many such pieces, with shared interior faces canceling since each is traversed with opposite outward normals from the two adjoining pieces). It suffices to prove the three component identities separately and add them; take the z-component. Write E=\{(x,y,z): (x,y)\in R,\ h_1(x,y)\le z\le h_2(x,y)\}. Then \iiint_E \frac{\partial R}{\partial z}\, dV = \iint_R \big[R(x,y,h_2(x,y)) - R(x,y,h_1(x,y))\big]\, dA by the single-variable Fundamental Theorem applied to the inner integral. The right side is exactly the flux of R\,\mathbf{k} through S's top piece z=h_2(x,y) (outward normal has positive z-component) plus its bottom piece z=h_1(x,y) (outward normal has negative z-component, contributing the minus sign) — the side faces of E, if any, are vertical and contribute nothing to a \mathbf{k}-component flux. The same argument for the other two components, summed, gives the theorem.